A beginer's qustion

Xiaoming Qian

Hi everybody. I'm developing an IC card handset device application with standard C language.I'm a new comer to C language.I have a question Look at the funtion definition : unsigned short dt_recv(unsigned char *data, int *datalen) Input parameters: Data : to receive data buffer Output parameters: Datalen:receive data buffer size Return values: none. Communication Protocol : data[0]=command head; ... data[3]=high of length; data[4]=low of length; ... Data[3] and Data[4] store a Int16 value.I have problem when I try to get the value.Here is my code: short i; memcpy(&Data[3],&i,2); But it seems that I did not get the right value.I guess I'm in the wrong way.I'm not familiar with memory operation. Can any one give me the right way?Any idea will be appropriate.

xyecloudy

are you a chinese boy?:laugh:

kakan

My suggestion (not tested): short i; i = (short) ((data[3] << 8) | data[4]);

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David Crow

Xiaoming Qian wrote:

short i; memcpy(&Data[3],&i,2); But it seems that I did not get the right value.I guess I'm in the wrong way.I'm

Are you wanting the value of Data[3] or i? The signature for memcpy() is:

void *memcpy( void *dest, const void *src, size_t count );

So what you have is writing from i to Data[3].

"A good athlete is the result of a good and worthy opponent." - David Crow

"To have a respect for ourselves guides our morals; to have deference for others governs our manners." - Laurence Sterne

Mark Salsbery

Xiaoming Qian wrote:

short i; memcpy(&Data[3],&i,2);

On Intel machines, i is stored with the least-significant byte first. If you use memcpy like this then Data[3] == least-significant byte of i Data[4] == most-significant byte of i To get the value back. something like: short i = (short)((unsigned short)Data[4] << 8) | (unsigned short)Data[3]; or short i; memcpy(&i,&Data[3],2);

"Posting a VB.NET question in the C++ forum will end in tears." Chris Maunder

Shatak Gurukar

Here "data[]" is an array. so "data" itself an a address. And instead of using "&Data[3]" use "Data[3]" in "memcpy" function That is: short i; memcpy(Data[3],&i,2);

Erik Midtskogen

I believe you should use a lower-case "D" in "Data". C is a case-sensitive language. For example, use memcpy(&i,&data[3],2); instead of memcpy(&i,&Data[3],2); As others have noted, there could be other problems because you're reading from an array of unsigned chars into a short. -- modified at 14:06 Wednesday 20th June, 2007

David Crow

Wrong. The first argument to memcpy() needs to be an address, of which Data[3] is not.

"A good athlete is the result of a good and worthy opponent." - David Crow

"To have a respect for ourselves guides our morals; to have deference for others governs our manners." - Laurence Sterne

David Crow

Erik Midtskogen wrote:

I believe you should use a lower-case "D" in "Data".

Umm, the compiler would have told him this (i.e., syntax error). :rolleyes:

"A good athlete is the result of a good and worthy opponent." - David Crow

"To have a respect for ourselves guides our morals; to have deference for others governs our manners." - Laurence Sterne

Erik Midtskogen

I was assuming that he may have been getting a compiler error and not known why. Probably we should ask him what sort compiler message or other symptom of a problem he is seeing.